Translation transformation
Case 1 as a function of \ (\ y = f (x ) = 2 ^ {| | x-1}) of the image.
Method: We choose \ (y = 2 ^ x \ ) based image transformation,
①先由\(y=2^x\xrightarrow{f(x)\rightarrow f(|x|)}y=2^{|x|}\)，
② Then a \ (y = 2 ^ {| x |} \ xrightarrow {f (x) \ rightarrow f (x-1)} y = 2 ^ {| x-1 |} \)
Symmetry transformation
① \ (Y = F (x) \ {xrightarrow symmetrical about the x-axis Y = -f} (x) \) ;
② \ (y = F (X) \ {xrightarrow symmetric about the y-axis y = F} (the -X-) \) ;
③ \ (Y = F (X) \ {xrightarrow symmetric about the origin Y = -f} (the -X-) \) ;
④ \ (Y = X ^ A (A> 0 and a \ neq 1) \ xrightarrow {symmetrical about the y = x} y = log_ax (a> 0 and A \ NEQ. 1) \) ;
Scaling
① \ (Y = F (X) \) \ (\ xrightarrow [when 0 <a <1, the abscissa is the elongation of the original \ frac {1} {a} times, the same ordinate] {if a> 1, the abscissa is the shortened original \ frac {1} {a} times, the same ordinate} \) \ (Y = F (AX) \) ;
② \ (Y = F (X) \) \ (\ xrightarrow [when 0 <a <1, the ordinate reduced to a times the original, unchanged abscissa] {if a> 1, the ordinate elongation a times the original, unchanged abscissa} \) \ (Y = AF (X) \) ;
Folding transformation
① \ (Y = F (x) \) \ (\ xrightarrow [the folded up below the x-axis image] {x axis side image retention} \) \ (Y = | F (x) | \) ;
② \ (y = F (X) \) \ (\ xrightarrow [symmetric about the y-axis image] {reserved right y-axis images, and for which} \) \ (y = F (| X |) \) ;
③ \ (y = F (X) \) \ (\ xrightarrow [symmetric about the y-axis image] {Reserved left image y-axis, and for which} \) \ (y = F (- | X |) \) ;